Problem: The grades on a geometry midterm at Springer are normally distributed with $\mu = 66$ and $\sigma = 3.5$. Brandon earned a $71$ on the exam. Find the z-score for Brandon's exam grade. Round to two decimal places.
A z-score is defined as the number of standard deviations a specific point is away from the mean We can calculate the z-score for Brandon's exam grade by subtracting the mean $(\mu)$ from his grade and then dividing by the standard deviation $(\sigma)$ $ { z = \dfrac{x - {\mu}}{{\sigma}}} $ $ { z = \dfrac{71 - {66}}{{3.5}}} $ ${ z \approx 1.43}$ The z-score is $1.43$. In other words, Brandon's score was $1.43$ standard deviations above the mean.